


MEADE 

4QA 

493 

Copy 1 


The New Mathematics . 

DAVID’S 


• i V upid Computing jsjjjgteni, 


With Key and Explanations for Home Study. 



Embracing a Collection of Original, Practical and Simple 
Methods of Calculating applicabje to every line of Busi¬ 
ness, and for the use of all who desire to become 
Quick and Accurate Accountants. 


AUTHOR OF 

David’s Interest Time Tables, and David’s Almanac for all Time. 

226 and 22S LAKE ST., CHICAGO. 


J. I>. Hki.a.v A Co., Prixtkks, Chicago. 

































The John J. and Hanna M. McManus 
and Morris N. and Chesley V. Young 
Collection 









\ 



I 






Copyright 1883 by Wm. K. David. All rights reserved. 

(?W Book pirates and brainless imitators beware: I will prosecute 
infringements to the fullest extent of my purse and the law. 








4-04 

493 


NEW LIGHTNING CALCULATOR. 


Lightning Addition. 

Wonderful rapidity and accuracy in addition are the result of 
thoroughly mastering the following table, which has been pre¬ 
pared with great care. It contains all the possible combinations 
of the nine digits in groups of three figures. Never add a single 
figure at a time. Every figure must "be seen and used, but not 
pronounced. There is as much reason for spelling each word in 
conversation as to pronounce each figure of an example while 
adding. 


Combination Addition Table. 

In memorizing keep your mind free from words, and do not 
say 9 plus 8 plus 7 equals 24, but simply nine , eight , seven , 
twenty-four. 


Combination of 3. 

Combination of 4. 

Combinations of 5. 

Combinations of 6. 

1 

1 

11 

242 

1 

1 

12 ‘ 

213 

1 

2 

32 

211 


Combinations of 7. 

Combinations of 8. 

Combinations of 9. 

Combinations of 10. 

3245 

12322 

123242 

12243236 

1211 

11123 

111213 

11212431 

3321 

65443 

765544 

87655443 


Combinations of 11. 

Combinations of 12. 

Combinations of 13. 

Combinations of 14. 

1232354363 

132435433 

23435436543 

1245456542544 

1112212314 

212121234 

21121231235 

4311211236345 

9877655544 

988776665 

99887776665 

9998887776665 


























Combinations of 15. 

Combinations of 16. 

Combinations of 17. 

Combinations of 18. 

3456547654645 

654765476465 

765487647656 

87658765766 

3211231234355 

123123423545 

123412353455 

12342345456 

9998887777665 

999888877766 

999988887776 

99998888776 


Combinations of 19. 


9876587676 

1234534556 

9999988877 


Combinations of 20. 


98768767 

23454566 

99998887 


Combinations of 21, 


9476877 

3856567 

9999887 


Combinations of 22. 


98787 

45667 

99988 


Combinations 
of 23. 

Combinations 
of 24. 

Combinations 
of 25. 

Combinations 
of 26. 

Combinations 
of 27. 

9878 

988 

98 

8 

9 

5678 

678 

78 

9 

9 

9997 

998 

99 

9 

9 


Lightning Combination Method and Practical 

Proof of Addition. 

Commence at the bottom of the right hand column. 97 
By grouping according to the table, three figures at a time, 976 
we produce the following combinations. They must be 465 
added as they occur. 234 

321 

9 + 16 + 12 + 12 + 8 + 15 = 72 562 

475 

Write the ten figure (7) over the ten column, and the 839 
"nit figure in its proper place at the foot of the unit col- 521 
umn. The next column contains the following combina- 642 
tions: 375 

211 

8 + 22 + 13 + 9 + 15 + 16 + 7=90 756 

483 

Write the 9 over the hundred column and the 0 under 296 
the ten column. Proceed in like manner with the remain- 757 
ing column, and write beneath the full result 9702. By 421 
writing and carrying figures at the top we have the ad van- 653 
tage of leaving our work in the middle of an operation, 715 

returning at leisure to finish the work without retracing - 

the figures already added. Only part of the answer of 02 
this example is written, so as to illustrate the beauty of 
this advantage. 




? /> 


0 

-> 

) 

.) 































Proof .—Commence at the left hand column, add down¬ 
ward the columns as they appear above. Set down the 9 7 
sum of each column as shown on the right, and cut off the 9) 0 
left hand figures, except the top one, and the answer is on 7) 2 
the right. 


Lightning Multiplication. 

By the following valuable rules and suggestions, the product 
of any two numbers can be obtained in a single line, without 
writing the partial products as by the ordinary method. 

The diligent student can master the art in a very short time, 
and thus secure rapidity and accuracy that cannot be surpassed. 

Reverse Process. 

Multiply 2341 by 522. 

Outline of Operation. 

F E D' C B A 
2341 2341 2341 2341 2341 2341 

225 225 225 225 225 225 

--- - -- Ans. 1222002 

12 2 2 0 0 2 . 

Explanation .—The above is an outline of the entire opera¬ 
tion. After the various positions are fixed in the mind, only one 
writing of the numbers is necessary. To learn this principle, 
take a slip of paper, and on the upper margin write the multi¬ 
plier in reversed order. (Thus 522 in reversed order reads 225.) 
Write the multiplicand on a separate paper. Place the left 
hand figure of the multiplier (2) under the right hand figure of 
the multiplicand (1), say 2x1=2, and write the product be¬ 
neath, as in the figures under A. Slide the paper one place to 
the left, and we say 2x1=2,2x4=8,8+2=10. Writedown 
the 0 and carry the ten. It will thus be observed that we handle 
only the products of the figures in the multiplier that come di¬ 
rectly beneath those in the multiplicand. Slide the paper one 
place again, and we find the figures in the position they occupy 
under O. We get the following products: 5x1=5, 2x4=8, 
2x3=6, 5 + 8 + 6 = 19. 19 + 1 (the carrying figure) equals 20. 
Write down the cipher and carry 2. Slide again, and we have the 
following: 5x4, 2x3, and 2x2=30 + 2, to carry, equals 32. 
Write down the 2 and carry 3. Slide again, and we have 5x3 and 
2x2, which, added, equals 19+3=22. Write down 2 and carry 
2. Slide again, and we have 5 x2=10, 10 + 2 = 12, which, when 









written beneath, completes the product. This process will mul¬ 
tiply all whole and fractional numbers. Study it carefully, and 
you wont regret it. 


Simultaneous Multiplication. 

A Practical General Pule. 

Multiply 23 by 52. 

Explanation. —We first say 3x2=6 (always reserve carrying 
figure, if any). We next say 3x5=15, and 2x2=4. 52 

15+4=19. Write down 9 and carry the 1. • 2x5=10. 23 

10 + 1=11. Write it down and the product is complete. - 

Ans. 1190 

Multiply 4532 by 71. 

Explanation. —First we say 1x2=2, and place it beneath. 
Next, 1x3 and 7 x 2=17. Write down the 7. 1x5 and 

7 x 3=26. 26 + 1, to carry,equals 27. Writedown the 7. 4532 

Next, 1x4 and 7x5 = 39. 39 + 2, to carry, equals 41. 71 

Write down the 1. Next, 7 x4=28. 28 + 4, to carry, -- 

equals 32, and completes the product. Ans. 321772 

Multiply 726 by 453. 

We first say 3 x 6 = 18, and write the 8 beneath. Next, 3x2 
and 5x6=36. 36 + 1, to carry, equals 37. Write down the 7. 

We next say 3x7, 4x6, and 5x2=55. 55+3=58. 

Write dowm the 8. Next 5x7 and 4x2=43. 43+ 720 

5=48. Write down the 8. To complete the product 453 

we say 4x7=28. 28 + 4=32, and write down the full- 

result. Ans. 328878 

Multiply 7854 by 6213. 

First we multiply 3 by 4, and get 12. Write down the 2. 
Next, 3x5 and 1 x4=19. Write down the 9. Next, 3x8, 2x4, 
and 1x5=37. 37+ 2, to carry, equals 39. Write down the 9. 

Next, 3 x 7, 6 x 4,1 x 8, and 2 x 5=63. 63 + 3, to carry, equals 66. 
Write down the 6 units. Next 1x7, 6x5, and 2 x 

8 = 53. 53 + 6, to carry, equals 59. Write down the 7854 

9. Next, 2x7 and 6 x 8 = 62. 62 + 5, to carry, equals 6213 

67. Write down the 7 and 6 x 7=42 + 0=18, write-- 

it down, and we have the answer. Ans. 48796902 

At first the answers will be written rather slowly, until the 
student is familiar with each step of the operation. The 
greatest objection to this rule for the beginner has been its 
retention of large numbers in the head. This is entirely over¬ 
come by studying the addition table in the first part of this work. 
To avoid forgetting the carrying figure, always add it in with the 





first product as you proceed. Surprising speed can be made with 
^his vule by any one who will learn and use it. 


Valuable and Rapid Contractions. 

To multiply any two figures by 11, as, for instance, 52 by 
11, we simply add the 5 and 2. We get 7, which placed between 
the 5 and 2 we have 572. Ans. If the figures when added pro¬ 
duce a sum greater than 9, write its unit tigure as above, and in¬ 
crease the tens figure of the multiplier by the 1 to carry. 

When the tens figures of any two numbers less than 
100 are alike, and their units add exactly 10, as in the 
following example, we simply write down the product of 96 

the unit figures and add one to the upper or lower tens 94 

figure we get 10, and multiply 10 by the other tens figure - 

we get 90, which prefixed to 24=9024. Ans. 9024 

This principle not only applies to numbers whose tens are 
alike, but can be extended to all numbers whose tens are dissimi¬ 
lar and whose unit figures add ten—as in the following. 

Multiply 68 by 42. 

First say 2 x 8 = 16. Write down the 6. Next subtract 
4 from 6, we get 2. 2x2 (the unit tigure of the smaller 68 

number)=4. 4 + 1, to carry, equals 5. Write it down and 42 

adding one to the tens figure of the larger number we get - 

7. 7 x 4=28, which prefixed to 56=2856. Ans. 2856 

To multiply any number of nines by any number in a single 
line by subtraction. 

Multiply 67423895647 by 99999999999. 99999999999 

67423895646 


A?is. 6742389564632576101353 

Explanation .—Write down the multiplier, less one , and sub¬ 
tract the multiplier, less one , from the number of nines. 

To multiply any number of miscellaneous figures by any 
number of like figures , as for instance, 3123 by 111, 222, 5555, 
1111, etc. 

Multiply 4254 by 3333. 

Explanation. —By this method we add 4. 6. 11. 15. 11. 9. 
horizontally the dissimilar figures in groups 4254 

according to the number of like figures, 3333 

and produce our product by multiplying - 

the groups by only one of the like figures. Ans. 14178582 

We first say 3x4=12, and write down the 2. We now 
commence grouping. 4 + 5=9, 9x3 (one of the like figures) 
equals 27+1, to carry, equals 28. Write down the 8. Next group 






4 + 5-1-2=11, 11x3=33, 33 + 2=35. Writedown the 5. Next 
we say 4 + 5+2+4=15, 15x3=45, 45+3, to carry, equals 
48. Write down the 8. Now the student will observe that we 
have arrived in our groupings over the last figure of our like 
figures. We now commence dropping our dissimilar figures from 
the right, one by one, as we group. Our next group will be 5 + 2 
+ 4=11, 11 x3=33 x4, to carry, equals 37. Write down the 7. 
Next we say 2 + 4=0, 0x3 = 18 + 3=21. Write down the 1. 
To complete the product we say 3x4=12 + 2, to carry, equals- 
14. Write down the full result. This last contraction is a 
grand improvement, and will rank high among the most impor¬ 
tant short rules ever discovered. The renowned accountant and 
expert, Mr. McMurray, of the Treasury Department at Washing¬ 
ton, uses it with great success. So far as the author’s knowledge 
extends, it has never before appeared in a mathematical work. 


Fractions. 

How to Add Fractions. 

Find the sum of f and 

Explanation. —First we say 7 x 5=35. Next we say 11 x 3 = 
33. 35 + 33 = 68, which is the numerator of our answer, and the 

product of the denominators, 7 x 11=77, the denominator of the 
answer. In actual business, the partial sums 
need not be written separately, but added men- M. 11 
tally as they occur. " i + T V=4f A ns. 

Find the sum of J, and -f. 

Explanation. —We lirst multiply 2 by 7 of the denominators j 
by 4 of the numerators; we get 56. Next, 2x5 (=denominators) 
equals 10. 10x3 (numerator) equals 30. Next w r e say 7x5 (de¬ 
nominators) equals 35. 35 xl (numerator) equals 35. 30 + 56 + 

35 = 121, the numerator of the answer. Multiply the denomina¬ 
tors, 7 X 5 x 2=70, the denominator of 
the answer. Proceed in like manner -21 lit lo. 
with any number of fractions. i. + 4 + 3 — Arts. 

How to Subtract Fractions. 

From f take f. 

Explanation.—3x3=9. 4*2=8. 8 from 9 leaves 1, the 

numerator of remainder. The product of the 
denominators 4 x 3=the denominator of the 1 1 
remainder, and we have y 1 ^. t Ans. 



Division of Fractions 

Rule. —Draw two parallel lines, and write the dividends on 
the upper line, and divisors on the lower line. Multiply the 
numbers outside the lines for the numerator, and those on the 
inside for the denominator of the answer. Always cancel factors 
common to both terms. 

Divide $ of § by f of $. l 2 ) 

3 4 | f = A Ans. 

~5~2 J 

By this ingenious arrangement, the divisors are inverted 
without the confusion and error so common to beginners. It is 
regarded by teachers and accountants, wherever introduced, as 
being the right method. A fraction in proper form represents a 
part of a unit, and, when inverted, represents the number of times 
it is contained in a unit. This is the reason for inverting the 
divisor in dividing fractions. 


Division of Mixed Numbers. 

To divide a mixed number by a whole number. 

Example .—Divide 9897-$- by 12. 

Explanation .—First find the number of times the divisor (12) 
is contained in the whole number of the dividend (9897), equals 
824, and a remainder of 9. Multiply this remainder by the de¬ 
nominator of the fraction of dividend, and add its numerator; 
thus, 9 x 8 = 72+ 7=79. This is the numerator of 
the answer. Multiplying 8 (the denominator Of the 12)9897$ 

dividend) by 12 (the divisor), we get 96, the denomi- - 

nator of the answer, and completes the result. Ans, 824$$ 


Multiplication of Mixed Numbers. 

General Rule . 

To multiply any two mixed numbers. 

Find the product of the whole numbers and the product of 
the fractions, and add to this amount the product of the lower 
fraction by the upper whole number, and the upper fraction by 
the lower whole number, and the product is complete. 



Multiply 21| by 15^. 

Explanation. —An outline of the entire operation 21^- 

is here given for tiie beginner. After the student is 15*- 

f am ilia/with the process, the partial products should —— 

not be written down, but added mentally as they occur. 315 
First we multiply 15 by 21, equals 315. Next we say 
| X 21=3, and x 15=3. Write them both down under 3 
315, also the product of \ by ^ equals -g 3 j . The sum is 0^ 

32 I 3 V Practice will give rapid results by this process.- 

Ans. . 3213^ 

Valuable Contractions in Multiplying Mixed Numbers. 

Practical Business Method. 

In actual business it is only necessary to get the answer to 
the nearest cent, as we have no coin representing less than that 
amount. When it is less than one-lialf a cent we drop it, and 
when one-half or more we call it one cent. The following simple 
rule of getting the answer to the nearest cent will be of great 
value to all, and should be learned by every one whose time is 
money. The entire disregard of fractions in the partial products 
renders a mistake or error almost impossible. 

Multiply 11\ by 9^. 

Explanation— Jx 11 = 3f. We call it 4, because llj 

it is nearer 4 than 3. £x9 = 2J, call it 2. 4 + 2=6, 9-J- 

plus the product of the whole numbers, 99, equals - 

105. Ans. 105 

Multiply 18f by 16-J-. 

Explanation. —18 x |-=4 to the nearest unit. 16 x j- 18J 

= 12. 4 + 12=16. The product of the whole num- 16-J- 

bers is 288. 288 + 16=304. - 

Ans. 304 

A valuable contraction for multiplying all mixed numbers 
whose fractions add one. 

Multiply 19J by 17J. 19 — 17=2. 2xf, the fraction of the 

smaller number, equals -J-. i+xV (the product of the 
fractions of the example) equals the fraction of the 19f 
answer. Now add one to the whole number of the mul- 17J 

tiplicand, we get 20. 20 x 17=340, the whole number --- 

of the answer. Ans. 340-^J. 

Multiply 89|- by 73f. 

Explanation. —Subtract 73 from 89 we get 16. 16 89f 

X f (the fraction of the multiplier) equals - 4 ¥ 8 - or 6. Now 731- 

add 1 to 89 we get 90 90 x 73=6570 + 6 = 6576. Then - 

add the product of the fractions fxf=-|-|-. Ans. 6576^|- 







From this extended illustration of the principle, we are satis 
tied the student will understand it thoroughly. The idea W6 
believe to be entirely original, and, as it covers such a wide rano-e 
ot numbers, we trust it will become generally known. 


Multiplication and Division of Aliquot Parts. 


The aliquot parts of 100 and 1000 enter so largely into the 
practical calculations of every-day business, that we here present 
a comprehensive explanation of those most generally used in busi¬ 
ness. 


33$ 

is 

i 

part 

of 

100 

333$ 

is 

i } 

part 

; of 

: 1000 

66$ 

u 

t 

66 

66 

100 

666$ 

66 

t 

66 

66 

1000 

25 

66 

i 

66 

66 

100 

250 

66 

$ 

66 

66 

1000 

75 

66 

i 

66 

66 

100 

750 

66 

4 

66 

66 

1000 

16$ 

66 


66 

66 

100 

166$ 

66 

£ 

66 

66 

1000 

12$ 

66 

1 

66 

66 

100 

125 

66 

i 

66 

66 

1000 

37$ 

66 

t 

66 

66 

100 

375 

66 

f 

66 

66 

1000 

62$ 

66 

f 

66 

66- 

100 

625 

66 

t 

66 

66 

1000 

87$ 

66 

i 

66 

66 

100 

875 

66 

f 

66 

66 

1000 


66 

tV 

66 

66 

100 

83$ 

66 

* 

6i 

66 

1000 

6$ 

66 

tV 

66 

66 

100 

62J 

66 

tV 

66 

66 

1000 

18$ 

66 

i 3 tr 

66 

66 

100 

187$ 

66 


66 

66 

1000 

31 $ 

66 

iV 

66 

66 

100 

312$ 

66 

T 5 2 

66 

66 

1000 


To multiply by an aliquot part of 100 or 1000. 

Pule. —Annex two ciphers to the multiplicand for 100 and 
three ciphers for 1000, then take such part of it as the multiplier 
is part of 100 or 1000. 

Multiply 9675 by 166f. 

Explanation. —166$ is $ part of 1000. Therefore annexing 
three ciphers to 9675 we get 9675000, or 1000 times 
that number. Divide this by 6 and we get $ of 1000 6)9675000 

times the number, or 166$ times the number. Ans. 1612500 


Multiply 56875 by 12$ 

Explanation. —12$ is $ part of 100, so wesim- 8)5687500 



How to divide by aliquot parts. 

Pule. —Pemove the decimal point of the dividend two places 
to the left for 100, and three places to the left for 1000, and 
multiply by such part as the divisor is part of 100 or 1000. 






Divide 9650 by 25. 

Explanation .—The decimal point is always on the right of 
whole numbers. By removing it two places on 9650 we 


get 96.50, or ^ part of 9650. 25 is \ of yfo, so it 96.50 

is contained 4 times 96.50, or 3S6 times. 4 


Ans. 386.00 

Divide 6740 by 12£. 

Explanation. —12^- is •§• part of Simply re- 67.40 

move the point two places for 100, and multiply by 8. 8 


We get 539A. The fraction is always in the form of a-- 

decimal. Ans. 539.20 


Interest. 


Decimal Rule for Days. —Year, 360 days per annum. 
One-hundredth part of the principal is the interest at 


7 <jo for 52 days.* 

7 U “ 48 “ 

8 i for 45 “ 

The above per cents, cover almost the whole range of interest 
'computations, and, as they are arranged, will be found easy to 


4 io for 90 days. 

5 i 0 “ 72 

<6 i “ 60 “ 


9 i for 40 days. 
12 i “ 30 “ 

10 i “ 36 “ 


memorize. 

Find the interest on $790 for 40 days at 9 per cent. 

Explanation .—Simply remove the decimal point two places 
to the left, which divides by 100, and we get $7.90 answer. The 
^student will observe that from this base to increase or diminish 
for other periods of time is an easy task. As, for instance, if 
we desire the interest for 50 days we add J, for 4 days, take 
for 400 days multiply by 10, etc. The reason for this rule is ob¬ 
vious. In one year a dollar at interest earns the number of cents 
represented by its per cent. At 9 per cent. 1 dollar in 1 
year earns nine cents, and in ^ of a year earns 1 cent. % of 360 
days equals 40 days. As 1 dollar earns 1 cent in 40 days, $7.90 
will earn 790 times $00.1 equals $7.90, ans. 

Required the interest on $814.50 for 56 days at 7J- per cent. 

Remove the decimal point two places to the left. 

We have $8,145, the interest for 48 days. We de- 6)$8.1450 
sire the interest for 56 days or 8 days more, hence 1.3575 

we add T 8 g- or + equals $9.50. - 

Ans. $9.5025 


This principle can be reversed, and the point removed on the 
days instead of the principal. The interest on $56 for 45 days 
at 8 per cent, is the same as the interest on $45 for 56 days- 


Nearly. 










Very frequently we can increase or diminish more readily by re¬ 
moving the point on the days instead of the dollars. How to take 
advantage of the.time when no advantage can be taken of the 
principal is one of the most important characteristics of com¬ 
puting interest. 

Rule. —Oue-lmndredth part of the days is the interest at 


4 ia on $90 

5 j* “ 72 

6 <f, “ ’ 60 


7 io on $52* 

7“ 58 

8 * “ 45 


9 f6 on $40 
10 io “ 36 

12 <f> “ 30 


Required the interest on $72 for 117 days at 5 per cent. 

Explanation .—Simply remove the decimal point two 
places to the left on 117 and we have $1.17. Ans . $1.17 

Required the interest oil $65 for 11 days at 7 percent. 

Explanation.— Remove the point two places 
on 11 we get 11 cents, the interest on $52. For 4).11 
11 days, $65 is $^f more than 52 or i of .11 .03 

=.03 to the nearest unit, which added to 11=- 

$0.14. Ans . .14 nearly. 


David’s Improved Six per Cent. Rule. 

Years, Months, and Days. 

Rule. —Multiply the years by 6, add one-half the number of 
months, and annex one-sixth of the number of days. Remove 
the decimal point three places to the left, and we have the inter¬ 
est on $1 for the given time; then multiply by the number of 
dollars, or dollars and cents in the principal. 

Example .—Riquired the interest on $341 for 5 years 8 
months and 24 days at 6 per cent. 

Solution. —5x6 = 30-1-one half the number of $341 
months equals 34. One-sixth of 24, the number of days, .344 

equals 4, which annexed to 34=344 or .344, the in- - 

tcrest on $1. .344 x $341=$117.304. Ans. $117,304 

When the month is of an odd number, as 7, 5, or 3, call the 
odd month 30 days, add it to the number of days, and then divide 
by 6 as before. This enables any one to instantly get the inter¬ 
est on $1 at 6 per cent, mentally, and, with the assistance of the 
rapid multiplication, only one line of figures is necessary to ar¬ 
rive at the most difficult interest calculation. 

When the days are not exactly divisible by 6, it is more con¬ 
venient to run them to one or two decimal places rather than use 
the fraction of sixths. 

v --- 

* Nearly. 







David’s Lightning Interest and Time Rule. 

Bases 1 and 10 per cent . 

A note for $40.80 is signed September 21, 1876, and paid on 
March 12, 1881, what is the simple interest at 10 per cent, per 
annum? 

Explanation .—Reject the century figures of the years for 
convenience. Write down 81 for the maturity year. March is 
the third month, therefore we write down 3, and 12 for the 
date of the month. Write down 76 for the year of signing, 9 for 
September, and 21 for the date of the month of signing. Our 
work will then appear as follows: 


Years. 

Months. 

Days. 

81 

3 

12 

76 

9 

21 


From 81 take 76=5 years, which, reduced to months, equals 60. 
60 + 3 months=63. Now annex of 12. In decimal form we 
have 63.4. Annex, in decimal form, -J of the 21 days to 9 (the 


number of months of the signing date). We have 
9.7. 9.7 from 63.4=53.7, or 53.7 months from Yrs. m os . Days, 

the date of signing to maturity. Now, to find the 81 63.4 12 

interest, multiply the number of months and tenths 76 9.7 21 

of a month by ^ of $40.80 (the principal). We get-- 

$3.40. 3.40x53.7=182.580. Take of this re- 53.7 
suit, and we have the interest at 1 per cent. Take 3.40 

■fa of it, and we have the interest at 10 per cent.=- 


$18,258. Ans. $18.2580 

If the interest is required for any other per cent., sim^y 
multiply the interest at one per cent, by the required per cent. 

When the days are not exactly divisible by 3, note the fo!low¬ 
ing: 

A note for $600, dated August 13, 1879, is due June 23, 
1880. Required the simple interest at 10 per cent. 

Explanation .—J une is the sixth month. The difference of the 
years I year or 12 months, plus 6 (the month of maturity) 
equals 18. of 23 (the date of the month of maturity) equals 7f, 
which annexed in decimal form to 18 = 18.7f. The month of 
signing was the eighth month, and 8 with -J of 13 (the date of 
signing) annexed = 8.4J. 8.4^ from 18.7f = 

10.333 (with fractions in decimal form). Al- 80 18.7-| 23 

ways extend the decimals to two or three places, 79 8.4J- 13 

or just far enough to produce the answer correct-- 

in cents. 10.333 + 50 (50 being of $600, the 10.333 
principal) equals 516.650. Remove the point 50 

one place and we have the interest at 10 per cent.,- 

equals $51.66+ans. Ans . $51.6650 






This is decidedly the shortest, simplest, and most general rule 
in existence. As far as the author’s knowledge extends, it is 
published in this form for the first time. 


Lightning Table for Marking Goods Bought by 
the Dozen. 

Retailers buy most of their articles by the dozen , such as 
boots, shoes, hats, caps, and notions of various kinds. A vast 
amount of time is employed in marking goods by the old pro¬ 
cess, and errors are frequent, owing to the unnecessary figures 
used. Purchases are sometimes made in a hurry, as, for instance, 
at auction, where goods frequently command a price per dozen 
that will not bear a living profit in the retail trade. If the pur¬ 
chaser will commit to memory the following table, he can in¬ 
stantly determine the retail price of a single article with any 
desired business per cent, added. The following per cents, are 
those usually used in business: 


To make 


20 #, divide the cost per dozen by 10 


33* 

a 

“ 9 




50 “ 

(4 

“ 8 




100 “ 


“ 6 




40 “ 


“ 10, 

and add 

i itself. 

35 

<4 

“ 10 

44 

i 

“ 

37* 

a 

“ 10 

44 

4 


30 “ 

<• 

“ 10 

44 

Vf 

«( 

25 

44 

“ 10 

44 

-fa 

u 

12* 

44 

“ 10, 

and subtract 

tV 

u 

16# 

44 

“ 10 

“ 

3+ 

u 

18* 

<4 

“ 10 

u 

TJ hs 

u 


W. J. L. Noel purchased one dozen hats, and marked them to 
sell at a profit of 33J per cent. What was the retail price of a 
single hat ? 

'Explanation .—According to table, simply divide 9) $27.00 

the cost per dozen by 9. - 

Ans. $3.00 

For how much must I sell books, bought at $28.00 per dozen, 
to gain 25 per cent? 

Explanation .—Removing the point one place to 24)2.800 
the left, on $28.00 we get $2,800. Now add -fa it- .11 

self, and we have $2.91+ ans. -- 

Ans . $2.91 + 

The table should be committed to memory. 










Discount. 


Discount is an allowance from the face of a bill. 

To deduct a specified per cent, from any number. 

Dule. —Multiply the number by $1.00, minus the rate per 
cent. 

Example .—A manufacturer offers to the trade aline of goods 
at a discount of 40 per cent, from the list price. What is the 
net value of a bill amounting to $675 \ 

Explanation .—40 per cent, discount, or $675 
t 4 o°o, from $1.00 leaves T W> or 60 cents per .60=1.00—.40 

dollar to be paid. - 

Ans. $405.00 

If sixty cents pays one dollar of the debt, .60 x 675 will pay 
die entire debt. 

When a discount is given for payment in a specified time, and 
only part of the bill is paid within limits of the time, employ the 
following rule : Divide the amount of payment by $1.00, minus 
the given rate. The result will be the amount to credit the ac¬ 
count. 

Example. —J. W. Dusenbury bought of J. A. Galvin & Co., 
Quincy, Mass., Mdse., $100 ; time 3 months, 6 per cent, dis¬ 
count. What amount should Dusenbury receive credit for2 

Operation. —1.00—.0.6=.94. 47.00-=-.94=$50, Ans. 


David’s Perpetual Mental Calendar. 

An accurate scientific method of telling mentally the day of 
the week from the commencement of the Christian era to time 
perpetual. An important rule for the student, accountant, and 
professional man. 


Monthly Excess Figures. 


0 

1 

2 

3 

4 

5 

6 

June. 

September 

December. 

April. 

July. 

* January. 
October. 

May. 

August. 

^February. 

March. 

November. 


* January and February are each called one less in leap years. 


















The excess figure to add for each month will be found above 
each name. 

Rule.— To the given year, omitting century figures, annex a 
cipher and divide by 8, rejecting fractions (if they occur). To 
this quotient add the date of the given month, its excess figure, 
and the excess figure of the century, and divide the amount by 7. 
The excess of sevens indicates the day of the week, as follows: 
A remainder of 1 indicates Sunday, 2 Monday, etc. An excess 
of 0 indicates Saturdav. 

Note. —No attention need be paid to the excess figure of this century, as it 
is 0. For the past century aud 2, and the coming century add 5. See Century 
Kules for other Centuries. 

Example 1.—Required the day of the week of July 4, 1876. 

Solution .—Omit 18, the century figures, and annex a cipher 
to 76, equals 760. 760+8=95 + 4 (the date of the month) 

plus 2 (the excess figure of July) equals 101. 101+7=14, with 

a remainder of 3, indicating the third day of the w r eek (Tues¬ 
day). If you desire the day of the w r eek of July 4,1776 or 1976 r 
simply add the century excess figure of the given century to 3, 
the above excess, and divide by 7 as before, if the sum of the two 
figures add 7 or more. When a date occurs in January or Feb¬ 
ruary in a leap year, be careful and call the monthly excess figure 
one less . 

Reason. —There are 52 weeks and one day in a common year. 
A leap year has one day more. Years exactly divisible by 8 are 
leap years. Century years exactly divisible by 8 are also leap 
years. We call the figures obtained after discarding the last two 
figures of the years, century years. In the year 2000 the century 
years are 20 and 20; being exactly divisible by 8 it is a century 
leap year. It is an error to suppose that every fourth year is a 
leap year. 

By dividing the year .8 we find the number of odd days since 
the beginning of the century. We add 3 for January, because 
January 1 , a.d. 1 , was on Friday. We add 6 for February, 
because January contains 4 weeks and 3 days excels, and this 
excess added to its own excess equals 6, the total number of 
odd days. This reason applies to all other months. We divide 
by 7 because there are 7 days in a -week. Lightning rapidity 
in answering questions by this rule may be attained by rejecting 
the sevens as fast as they occur. The excess figures of the 
months should be memorized. 

In the year 1583, in all Catholic countries the Julian calendar 
or old style was dispensed with, and the Gregorian calendar or 
new style adopted. The new style was adopted in England in 
1752. The old style is used in Russia still. 


Century Rules. 


How to find the excess figure of any century from lbO to 
1500, inclusive. 

Rule. —Reject the last two figures of the year and multiply by 
6 . To the product add 2 and divide by 7. The remainder of 
sevenths is the excess figure to add for the century. 

Example .—Required the excess figure to add for dates occur¬ 
ring a.d. 1200 to the next century following. 

Operation and Solution .—Rejecting the last two figures of 
1200 we get 12. 12x6=72, 72 + 2=74, 74-^7=10, and a re¬ 

mainder of 4. Consequently 4 is the excess figure. 

How to find the excess figure of any century from 1600 to 
time perpetual. 

Rule. —Reject the last two figures of the year, and deduct 
16. Multiply the remainder by rejecting fractions. To this 
amount add 4 and divide by 7. The remainder is the excess 
figure. 

Example .—Required the century excess figure of a.d. 2400. 

Operation and Solution .—Rejecting the last two figures of 
2400, we get 24. 24-16=8, 8x5i=42, 42 + 4=46, 46^-7 

=6, and a remainder of 4. 4 is the excess figure. 

Examples for Practice .—The battle of Winchester was fought 
May 25, 1864; required the day of the week. Ans. Wednesday. 

Patrick Henry was born May 29, 1736; required the day of 
the week? Ans. Tuesday. 

William Shakspeare was born April 22,1564, and died on the 
anniversary of his birthday in 1616 ; required the day of the week 
of both dates? Ans. Born Saturday, died Friday. 

What day of the week were you born ? 


Practical Mensuration. 

Embracing many new and original ideas and short methods. 
Ar ranged for easy business reference, special classes in schools, 
and all who wish to become skilled in this important branch of 
mathematics. 


Cylindrical Capacity. 

To find the cylindrical contents of a cylinder. 

Rule.— Square the diameter and multiply by the depth. 

To find the cylindrical contents of a frustum of a cone. 

Rule.— To the product of the upper and lower diameters add 
the sum of their squares, and multiply by one-third of the depth. 



To change cubic to cylindrical measure. 

Rule.— Divide the cylindrical contents by .7854. 

To change cylindrical to cubic measure. 

Rule. —Multiply the cylindrical contents by .7854. 

Note. —A cylindrical foot is the volume of a cylinder one foot in depth and 
diameter, and contains 1728 cylindrical inches. It is equal to .7854. or for greatez 
accuracy .785398 ■+- part of a cubic foot. A cylindrical foot contains 1357.16 + 
■cubic inches, aud a cubic foot contains 2200.15+ cylindrical inches. 

Cubio Capacity. 

To find the cubic contents of a cylinder. 

Rule.— Find the cylindrical contents by the preceding rule* 
and multiply by .7854; or, square the diameter in inches, multi¬ 
ply by the depth in feet and by .005454-h. 

To find the cubic contents of a frustum of a cone. 

Rule.— Find the cylindrical contents by the preceding rule, 
and multiply by .7854. 

To find the cubic contents of a cone or a pyramid, whether 
round, square, or triangular. 

Rule.—M ultiply the area of the base by one-third of the alti¬ 
tude. 


Measurement of Area. 

To find the diameter of a circle. 

Rule.— Multiply circumference by .31S31; or, extract square 
root of area and multiply by 1.12838. 

To find the circumference of a circle. 

Rule.— Multiply the diameter by 3^ or 3.1416 ; or, multiply 
radius by 6.283185. 

To find radius of a circle. 

Rule.—M ultiply circumference by .159155, or square root of 
area multiplied by .56419. 

To find the side of ail inscribed square of a circle. 

Rule.— Multiply diameter by .7071. Multiply circumference 
by .225. 

To find the side of ail equal square of a circle. 

Rule.— Multiply circumference by .282; or, multiply diam¬ 
eter .8862. 

To find the side of an inscribed equilateral triangle. 

Rule. —Multiply diameter by .86. 

To find the area of a circle. 

Rule. —Multiply the square of the radius by 3.1416; or, 
multiply the square of the diameter by .7854 ; or, square the cir¬ 
cumference and multiply by .07958; or, multiply one-half the 
circumference by half the diameter. 

To find the area of an ellipse. 

Rule. —Multiply both diameters together and by .7854. 



To find the area of an inscribed square of a circle. 

Rule. —Multiply the diameter by one-half itself. 

To find the area of a triangle. 

Rule. —Multiply base by one-half the altitude. 

To find the area of a trapezoid. 

Rule. —Multiply the altitude by one-half the sum of Its- 
parallel sides. 


Cistern, Tank, and Barrel Measure. 

Find the cylindrical feet, and for 
Gallons, multiply by 5.875 or 
Barrels “ “ .1865 

Hogsheads “ “ .09825 

Gallons “ “ 4.8126 

Or, find the cubic feet, and for 
Gallons, multiply by 7.48 

Barrels “ “ .2374 

Hogsheads “ “ .1187 

Gallons “ “ 6.1276 

Examples. 

IIow many barrels wine measure are contained in a cylir* 
drical cistern, the diameter 4 feet and the depth 6 feet. 

Operation .—Find the cylindrical feet by squaring the diam¬ 
eter and multiplying by the depth, which equals, 4x4=16, 16 x 
6=96. Now multiply 96 by .1865 (the part of a barrel in oim 
cylindrical foot) and we have 17.9 + barrels. 

How many gallons wine measure are contained in a railroad 
tank in the form of a frustum of a cone, the dimensions being as- 
follows* Height, 15 feet; lower diameter, 22 feet; upper diam¬ 
eter, 20 feet. 

Operation and solution .— 

400 =square of upper diameter. 

484 = u “ lower diameter. 

440 =prodnct of upper and lower diameters. 

1324 

5 =^- of the depth. 

6620 =cylindrical feet. 

5-J=capacity of one cylindrical foot. 


V Wine measure. 
Beer measure. 

| Wine measure. 
Beer measure. 


38892.5 =gallons. 






To find the number of gallons in barrels or casks. 

Rule.— Take all the dimensions in inches. Add the head 
and bung diameters and divide by two for the approximate mean 
diameter. Square the mean diameter and multiply by the depth. 
Multiply the result by .0034 for wine gallons, and .0028 for beer 
gallons. 

How many wine gallons are contained in a cask, the bung* 
diameter of which is 22 inches, the head diameter 24 inches, and 
the depth 30 inches. 

Operation. —22x24=46. 46-^2=23 (mean diameter). 23 2 = 
529. 529x30 (length) =15870. 15870 x .0034=53.9 + gallons.. 


Grain Measure. 

Rule.— Find the cubic feet, and for U. S. standard bushels 
multiply by .80356; imperial (English) bushels multiply by .779. 
Or 

Rule.— Find the cylindrical feet, and for U. S. standard 
bushels multiply by .63111; imperial (English) bushels multiply 
by .61183. 

Examples. 

A granary 10 feet long, 5 feet wide, and 8 feet high is filled 
with wheat. Required the number of U. S. standard bushels. 

Operation .—.80356 =capacity of one cubic foot. 

400=cubic feet. 


Ans. 321.4 + = bushels. 

Corn in the Ear. 

The number of bushels of shelled-corn in any space filled 
with corn on the cob can be only approximately determined, 
owing to various conditions affecting the kernel and cob. The 
following simple rules are, however, recognized by custom in 
nearly all sections of our country. 

Rule. —To find the number of bushels multiply cubic feet by 
9 and divide by 20, or multiply cubic feet by .45, or divide cubic 
inches by 3840. • 

To find the number of Tennessee barrels (5 bushels each). 

Rule. —Multiply cubic feet by .09. 

Examples. 

A crib 10 feet long, 8 feet high and 5 feet wide is filled with 
corn on the cob. How many bushels of shelled-corn does it con¬ 
tain. 




Operation. — 45 ^capacity of one cubic foot corn measure. 
400=cubic feet. 


Ans. 180.00 bushels. 


Log Measure. 

To find the number of feet, board measure , contained in any 
log. 

Kule. —From the diameter of the log at the small end in 
-inches , subtract 4. Square one-fourth of the remainder and 
multiply by the length in feet. 

To find the number of cubic feet in a round log in the shape 
of a frustum of a cone. 

Rule. —Square both diameters in inches , and to their sum 
add the product of the diameters. Multiply by the length in 
feet and by .001818. 

To find the number of cubic feet in the largest square piece 
of timber that can be sawn from a round log. 

Rule.— Multiply the diameter of the small end in inches by 
one-half itself and by the length in feet. Divide the result by 
144, or multiply the diameter in feet by one-half itself, and by 
the length. 

Examples. 

How many feet of inch boards can be sawn from a log the 
■diameter at the small end being 28 inches and the length 11 feet. 

Operation .— 28=diameter in inches. 

4 

4) 24 

Ans. 6 2 =36. 46x11=396. 

How many cubic feet are contained in a round log 30 feet in 
length, diameter of small end 10 inches, and large end 20 inches, 

Operation. —100=square of diameter, small end. 

400=square of diameter, large end. 

200 =prod net of small and large diameters. 

700 

30=lengtli in feet. 

Ans. 21000 x .001818 = 38.178 feet. 

How many cubic feet are contained in the square of a round 
log 18 feet long, and diameter of the small end 4 feet. 

Operation .—4-i-2=2=(-| of diameter) 4 (diameter) multi¬ 
plied by 2=8. 8 x 18 (length) = 144 cubic feet. 





Lumber Measure. 

Lumber is bought and sold by the foot, board measure , which 
is a square foot one inch in thickness. The standard lengths of 
planks, joists, scantlings, etc., are all the even numbers ranging 
from 8 to 30 feet. The following will be found to be the short¬ 
est cuts possible to produce correct answers: 


1 

inch thick. 

, 8 feet long, 

multiply width 

in inches by 

2 

3- 

1 

tt 

10 

tt 

tt 

a 

i 

1 

tt 

12 

tt 

it 

a 

0 

1 

1 

. tt 

14 

tt 

u 

a 

ill 

1 

tt 

16 

tt 

u 

a 

H 

1 

tt 

18 

tt 

a 

u 

0 

H 

1 

tt 

20 

tt 

a 

u 

if 

1 

a 

22 

a 

a 

a 

if 

1 

a 

24 

.< 

u 

a 

2 

1 

tt 

26 

tt 

a 

u 

2 i 

1 

tt 

28 

tt 

u 

tt 

H 

1 

" tt 

30 

a 

a 

tt 



General Rule.— Multiply the length in feet by the width 
and thickness in inches and divide by 12. 


Wood Measure. 

Wood is bought and sold by the cord. The length of a cord 
is 8 feet, the height 4 feet, and the width anything that may be 
agreed upon by buyer and seller. 

It is an error to suppose that a cord of wood should always 
contain 128 cubic feet, as, for instance, cooking-stove wood is 
seldom over 20 inches in length, and when corded the width of 
the pile is far from being 4 feet. 

Rule.— Multiply the length of the pile in feet by the height 
in feet, and divide by 32. 

How many cords of wood in a pile 17 feet long, 6 feet high, 
and 4f feet wide. 

Operation — 17=length. 

6=height. 

» • 

32 )102( 3 t 3 ^- cords Ans. 

96 


6 



Brick Measure. 

Bricks are of different sizes, varying according to custom of 
State or locality. They are estimated by the thousand. When laid 
in walls one-eighth is allowed for mortar. A wall one layer 
in thickness is called a 4^-inch wall; two layers, a 9*inch wall; 
three layers, a 13-inch wall ; four layers, an 18-inch wall; and 
five layers, a 22-inch wall. In most localities no deduction is 
made for openings, such as windows, doors, etc., as the extra 
work required is considered fully equal to plain work. 

General Rule.— Multiply the number of square feet in the 
wall or house by the number of layers in thickness and mul¬ 
tiply the product by the number of bricks shown in a square foot 
when laid in walls. 

The following convenient and simple table gives the names 
•of six standard bricks, together with their dimensions and the 
number of each contained in a surface, or square foot, of a 44- 
inch, 9-inch, 13-inch, 18-inch, and 22-inch wall. 


Name. 

Size. 

No. of 
bricks per 
surface 
foot 4%- 
inch wall. 

No. of 
bricks per 
surface 
foot 9-inch 
wall. 

No. of 
bricks per 
surface 
foot 13- 
inch wall 

No. of 
bricks per 
surface 
foot 18- 
inch wall. 

No. of 
bricks per 
surface 
foot 22- 
inch wall. 

Baltimore... . 

8ix4ix2f 

6 £ 

13 

m 

26 

32 i 

New York.. . 

8 x x 

n- 

14| 

21 i 

m 

35| 

Maine. 

^x3|x2| 

7.3 

14.4 

21.6 

28.8 

30 

Michigan .... 

8 i X 4£ X 2f 

6f 

12 f 

19 

25J 

31§ 

Fire . „. 

x 4f x 2f 

6 t 

13* 

20 

26f 

33* 

Common .... 

8x4x2 

8 

16 

24 

32 

40 


Example .—How many Baltimore bricks are contained in an 
18-inch wall 56 feet long and 25 feet high ? 

Solution and Operation .—56 x 25 = 1,400 square feet. Refer 
to Baltimore in the table, and directly opposite, in the 18-inch 
wall column, we find 26, the number of bricks per square foot, 
which, multiplied by 1,400, equals 36,400. 


Stone Measure. 

Stone when laid in walls, fences, foundations, etc., is usually 
-estimated by the perch, which contains 24f cubic feet. This 
mixed number is a very difficult divisor, and numerous attempts 
















liave been made to save reducing it to an improper fraction or 
■decimal. The following simple rule is the shortest known : 
Rule.— Find the cubic feet and multiply by .0404. 

A wall 100 feet long, 5 feet high, and two feet thick, contains 
how many perches. 

Operation .—.0404 = part of perch in one cubic foot. 

1000=cubic feet. 


40.4000=perches. 

The student no doubt observes that this is an approximation 
although sufficiently exact for business purposes. If greater ac- 
•curacy is required, add of the answer thus obtained. 


Carpenters’ Rule to Make Square Timber Octa¬ 
gon or Eight-sided. 

Place a 24-inch rule across each side of the timber, with the 
ends of the rule at the edges of the stick, and then point off 7 
inches from each end of the rule. Cut down to the points. 

If a 12-inch rule be used, point off 3£ inches from each end. 


Mechanics’ Iron Rule. 

To find the weight of wrought iron by measurement. 

Rule.—M ultiply the thickness and width in inches by one- 
third of ten times the length in feet. 

For—Cast iron, deduct 
Steel, add 

Copper, “ 

Cast brass, “ 

Lead, 

Example .—How many pounds of wrought iron are contained 
in a bar 15 feet long, 1J- inches thick, and 2 inches wide? 

Operation. —l£x2=3. Ten times the length equals 150. 
150-r-3=50, 50 x 3=150 pounds. 


TTT* 

A 

T* 

tV 










DAVID'S PERFECTED BUSINESS RULE 

FOR 

-^Multiplying all Decimal and Mixed Numbers,^ 


The lowest unit in United States money being the cent, it is useless to entail loss of time, and 
incur unnecessary liability to error, by seeking strict mathematical accuracy, when all that is- 
practically needed is correctness as far as cents, or the place of hundredths. 

By the following strictly practical and scientific rule, all unnecessary figures, as well as the 
mental labor expended in producing them, are obviated ; and the rule can even be used as a single 
line process, after the student becomes thoroughly familiar with every detail of the operation. 
The work is still further facilitated by the reduction of common fractions, in either factor, to their 
equivalent decimals, thus ensuring the simplicity attending operations in whole numbers. 

We will now proceed to illustrate the full application of the rule by the following: 


EXAMPLES. 


Example i. —Required, the product of 824.375X.005. 


OPERATION. 
PERFECTED PROCESS. 

824.375 

.005 


4.12-{- .4 ns. 


OPERATION. 
ORDINARY PROCESS. 

824.375 

.005 


4.121875 Ans. 


Explanation. —The work of multiplying invariably commences with the left hand digit of the 
multiplier, and the order or place filled by this digit determines the beginning figure of the multi¬ 
plicand. Though immaterial by which factor we multiply, it is advisable to use the factor containing 
the smaller whole number as the multiplier. In this example the multiplier consists of but one 
digit (5) occupying the thousandths place, and the digit occupying the thousandths place in the 
multiplier should never be used to multiply any figure on the right of the tens place in the multi¬ 
plicand, except to find mentally the carrying figure, which is found by multiplying the nearest 
discarded figure of the multiplicand and adding the tens of its product to the product obtained 
from the beginning figure of the multiplicand, increasing the said tens figure by the addition of one 
more ten whenever the units figure on its right is 5 or more. Sometimes, in order to ensure the 
greatest accuracy as to hundredths in the final result, it will be necessary to note, and make allow¬ 
ance for, the tens which would result from multiplying two of the figures on the right of the 
nearest discarded figure in the multiplicand, as is shown in Example 3. 

In accordance with the foregoing, we now say 5 times 2 (tens in the multiplicand) equals 10, and 
2 to carry from the multiplication of the nearest discarded figure (5X4=20) are 12. Set down the 
units figure (2) in the hundredths place, and carry the 1 ten to the product of 8X5, which makes 41; 
prefix this to the units figure, and we have 412 hundredths, the required answer, or 4.12 (four and 
twelve one hundredths). The answer is always hundredths, which may be reduced as required. 

Example 2. —Required, the product of 824.375X.015. 


OPERATION. 
PERFECTED PROCESS. 

824.375 

.015 


8.24 

4.12 

12.36-f- Ans. 


OPERATION. 
ORDINARY PROCESS* 

824.375 

.015 


4121875 

824375 


12.365625 Ans. 











Explanation.— In this example, the left hand digit of the multiplier is in the hundredths place; 
therefore, the work of multiplying must begin with the units figure of the multiplicand (4), and 
proceed to the left in succession, as by the old or ordinary method, except as to the carrying figure, 
to be obtained as has been explained above. Relatively to the digit of the multiplier (1), 3, 7, 5, of 
the multiplicand, are discarded figures. 

Now, multiplying the units of the multiplicand (4) by the digit in the hundredths place of the 
multiplier (1), we obtain the product 4, with no tens to carry, because the product of the nearest 
discarded figure (3), by the multiplier (1), is less than five. Set down the result (4) in the hun¬ 
dredths place, and proceed to the left as before, and, prefixing the result (82) to the figure in the 
hundredths place, the first partial product obtained is 824. Next, find the product of the multipli¬ 
cand by the digit in the thousandths place (5), as explained in the preceding example, to-wit, 412, 
and place it beneath the first partial product, so that hundredths will fall beneath hundredths, 
tenths beneath tenths, etc. It must be borne in mind that all partial products are to be set down 
in such a way that units of the same denomination will fall in the same column, hundredths under 
hundredths, tenths under tenths, units under units etc. Finally, adding all the partial products 
together, their sum will be the product sought. 


Example j. —Required, the product 

OPERATION. 
PERFECTED PROCESS. 

824.375 

.615 


49463 

824 

412 


506.994- A ns. 


824.375 X.615. 

OPERATION. 
ORDINARY PROCESS. 

824.375 

.615 


4121875 

824375 

4946250 


506.990625 Ans. 


Explanation. In this example, the first digit on the left, or the beginnimr figure of the multi¬ 
plier (6), occupies the tenths place; and attention may properly be called here to the most impor¬ 
tant fact, that, in this new and perfected process of multiplication, the digit occupying the tenths 
place in the multiplier is the, so to speak, pivotal figure—that is to say, it is the figure by which 
the beginning figures of the multiplicand are always and easily determined. 

The first digit of the multiplier (6) being in the tenths place, the beginning figure of the multipli¬ 
cand must always and invariable be that found in the tenths place. Hence, we say, 6 times 3 (tenths) 
are 18, and fiveto carry (from the two discarded figures of the multiplicand, 7 and 5,) are 23. There 
are 5 tens to carry, because inspection shows that 6 times 5, or 30 give three tenths to add to 6 times 
7, or 42, making 45; and, as here, the units figure (5) on the right of the tens (4) equals 5, one more 
ten must be added to the four tens, as accounted for in preceding explanation. Now, set down the 
units figure (3) in the hundredths place, as already directed, and carry the 2 (tens) to the product 
of the remaining figures of the multiplicand (824) by the multiplier, which makes 4946; and pre¬ 
fixing this result, as before, the first partial product becomes 49463 (hundredths). The partial pro¬ 
ducts by the remaining figures of the multiplier (1 and 5), will be obtained a3 set forth in examples 
1 and 2. 


Example U .—What is the product of 824.375 by 9.615? 


OPERATION. 
PERFECTED PROCESS. 

824.375 

9.615 


741938 

49463 

824 

412 


7926.37— A ns. 


OPERATION. 
ORDINARY PROCESS. 

824.375 

9.615 


4121875 

824375 

4946250 

7419375 


7926.365625 Ans. 


Explanation In this example, the first digit in the multiplier, on the left, is in the order of 
units- whenever this occurs, the multiplication will begin with the hundredths figure of the multi¬ 
plicand therefore, we say 9 times 7 are 63, plus 5, the carrying figure, (from 9 times 5, the 
nearest discarded figure) are 58. Place the units figure (8) in hundredths column, and carry the 












tens (G) to the product of the remaining figures of the multiplicand (82437), multiplied in their pro* 
per succession to the left, and the first partial product is 741938 (hundredths). Find a3 previously 
directed, the partial products of the remaining digits of the multiplier, setting down the first, or 
right hand figure of each, in the order of hundredths, and their sum will be the required product 
to the nearest hundredth, viz.: 7926.37. 

In the foregoing examples, it will be observed that the same quantity has been assumed as the 
multiplicand, while the multiplier, has been changed by the introduction of an additional digit, but 
preserving those already operated with. This has been done to facilitate the learner’s progress, by 
conducting him over ground already traveled, and to avoid perhaps tiresome repetition in the ex¬ 
planations. But the method illustrated by those examples will apply to any and all similar quanti¬ 
ties or numbers, no matter what or how many the digits composing them, with the same lessen¬ 
ing of labor and certainty of correct results. i . 

It may happen, and does most frequently in actual business, that one or more p aces in the 
multiplier may be filled with a cipher. In such case, how proceed? Simply pass on to the next 
significant figure of the multiplier, omitting each operation required were there a digit in the order 
or place filled by the cipher. For instance: 


Example j .—What is the cost of 36) yd3. of cloth at $1.07) per yard. 


OPERATION. 
PERFECTED PROCESS. 

36.50 

1.075 

3650 

256 

18 


$39.21- Ans. 


OPERATION. 
ORDINARY PROCESS. 

$1,075 

36.5 


5375 

6450 

3225 


$39.2375 


Explanation. In the Perfected Process, we write the number of j r ards in decimal form (36.5) for 
the multiplicand, and the price in decimal form for the multiplier, as it is the factor containing the 
smaller whole number. The left hand digit of the multiplier being in the place of units, the begin¬ 
ning figure of the multiplu and will be that in the hundredths place; so we must always annex a 
cipher if there be no digit i. \ the hundredths place, and the multiplicand, in this instance, is written 
36.50. Now, we say 1 time 3650 is 3650, and the next place in the multiplier being filled with a 
cipher we pass it, as, also its beginning figure in the multiplicand, and say 7 times 36 with 4 to 
carry, equals 256. Next, we have 5 times 3 plus 3 to carry, are 18; and, adding these partial pro¬ 
ducts together, we obtain the complete product 3924 (hundredths), 39.24. 

Again, it may occur that there are not digits enough in the multiplicanu to permit of multiplying 
by the extreme right hand figure of the multiplier, and the result would be slightly inaccurate, 
because of the absence of the last carrying figure. In such cases, the last carrying figure will be 
obtained by menta'ly prefixing a cipher to the multiplicand, and, to its product, by the multiplying- 
figure, adding the tens to be carried from the multiplication of the digit on the right of the prefixed 
cipher; the carrying figure thus procured will be the last partial product. 

In illustration, let it be required to find the product of 426.1875X.32456. 


OPERATION. 
PERFECTED PROCESS. 

426.1875 

.32456 


12786 

852 

170 

21 

3 


138.32-f ^A«s. 


OPERATION. 
ORDINARY PROCESS. 

426.1875 

.32456 


25571250 

21309375 

17047500 

8523750 

12785625 


138.323415000 jins. 


Explanation. —Having obtained, by the Perfected Process, the partial products, except the last 
(3). in the manner laid down in the foregoing explanations, there still remains the extreme right 
hand digit of the multiplier (6), to be used as a partial multiplier. The partial product (21) was had 
by multiplying the hundreds in the multiplicand (4) by the tenths of thousandths in the multiplier 











(5), leaving no beginning figure in the multiplicand for the partial multiplier in the hundredths of 
thousandths place (6), the beginning figure for such partial multiplier being thousands in the mul¬ 
tiplicand. Complete the multiplication by mentally prefixing a cipher to the multiplicand, and 
saying: 6 time 0 plus 3 to carry (from the discarded figures 2 and 6) are 3; which result set down 
in its proper place (hundredths), and thus the last partial product is secured. 

The problems just solved exemplifies in a manner that cannot fail to be remarked, even by the 
least observant, the wonderful superiority of the Perfected Process of Multiplication over the 
antiquated method of our schools. If it is true that, in these days, “time is money,” why waste 
our money, and overtask our mental powers to achieve, as by the ordinary method, a result, so 
large a portion of which is discarded as of no practical moment, after it has been so laboriously 
obtained 1 

It will be remarked, further, that while it is deemed preferable, in this New Process, to commence 
the work with the left hand digit of the multiplier, as has been done in all of the foregoing opera¬ 
tions, yet the right hand digit, or any other, in the factor by which we multiply, may equally well 
be taken as the first partial multiplier, care being had to observe the proper beginning figure of the 
multiplicand, as laid down in the explanations above give- 


We proceed, now, to embody the above directions in the following 
GENERAL RULES. 

r. Select , as the multiplicand, where both factors are decimals, that containing the fewest 
figures, or the mixed decimal containing the greater ivhole number, and, beneath it, place the 
other factor, so that units of the saine denomination will fall in the same column: units under 
units, tenths under tenths, hundredths under hundredths, etc. 

II. The work of multiplication, by the digits of the multiplier, begins with a certain place, or 
order, in the multiplicand, according to the place filled by the partial multiplier, and proceeds 
from that beginning place to the left, as by the ordinary method, each partial product obtained 
being increased by the tens to carry from the product of the discarded figure or figures nearest to 
the beginning figure of the multiplicand, by the partial multiplier. If the partial multiplier 
occupy the tens place, the beginning figure of the multiplicand is thousandths, if the units place, 
hundredths; if the thousandths place, tens; if the ten-thousandths place, hundreds; and so on, 
each digit further to the right in the multiplier controlling the next place to the left in the 
multiplicand. 

HI if a cipher occur in any order in the multiplier, pass to the next significant figure, and 
proceed as the latter requires; and, if there be not digits enough in the multiplicand to supply 
the beginning places controlled by the different figures of the multiplier, annex to the multiplicand 
as many ciphers as may be required. 

IV. Set down the right hand figure of .each partial product beneath the hundredths of both 
factors. 

V. Add all of the partial products together, and their sum will be the required product in the 
denomination of hundredths. 

jjote. _Removing the point from right to left in the multiplicand, and proceeding as above, will produce the 

answer to any period of exactness. 



TRADE DISCOUNTS. 


Trade Discounts are deductions allowed by manufacturers and jobbers from list prices. 

RULE. Multiply the list price by 100%, minus the rate %. 

Example 1 .—A manufacturer allows an agent a discount of 37^% from machinery, the list price 
■of which is $212.50. Required, the net price. 


OPERATION. 
ORDINARY METHOD. 


OPERATION. 
RAPID METHOD. 



$212.50 

.625 


12750 

425 

106 


106250 

148750 

63750 


$132.81 Ans. 


79.68750 


212.50 

79.69 


$132.81 Ans. 


Explanation. —37 % off, leaves 62£ on; or, in other words, 62^c. per dollar to be paid. Therefore, 
•we simply multiply .625 by the number of dollars. 

Frequently a series of rates, as 25%, 10% and 5% are used to produce the net price. According to 
the usual method, the first rate % of the series is deducted from the list price, and the others are 
deducted from the remainders in their regular order. It wi.l be found immaterial which is first 
used—60 and 10 off being the same as 10 and 60 off. 

The first object is to find the amount per dollar to be paid, and then multiply by the number of 
dollars, as before shown; hence we have the following: 

RULE. From 100 % deduct each rate % of the series separately, and multiply the product of all 
the remainders by the list price. 

Example 2 .—A merchant purchased a bill of hardware amounting to $763.50. A discount of 
.65%, 33£%, 25% and 20% was allowed. Required, the net result. 


OPERATION. 
RAPID METHOD. 


OPERATION. 

SHORTEST ORDINARY METHOD. 


$763.50 

.18 


$763.50 • 
.45 


7635 

6U8 


381750 

305400 


$137.43 Ans. 


$343.5750=55% off. 
2 


$229.0500=33J% off. 


.75 


11452500 

16033500 


$171.787500=25% off. 
80 


$137.43000000=20% off. Ans. 





















Ex p LAN ATr °\.—The rates, 55%, 33$%, 25% and 20%, deducted separately from 100%, leaves 45%, 
75% and 80%. The product of these remainders can be produced mentally with very little 
labor, thus: .45X .63§= .30, and .75X.80=.60. Now, simply say: .60X.30= .18%, the average 
rate per dollar of the list price to be paid. This, multiplied by $763.50=8137.43. Ans. 

By this method, it will be observed that salesmen and dealers can almost instantly ascertain the 
average or net rate. In working the example by what is termed the “Shortest Ordinary Method,” 
care has been used to make it appear as short as possible. A glance at the workings of the two 
methods will convince the student of the superiority of the Short Method. 


HSaoiTs ©ambirtoil Time and Interest Tables, 


RATES 6%, 7%, 8% 10% AND 12% PER ANNUM. ASSUMED YEAR 360 DAYS. 

Showing at a glance the Difference in Time between Dales and the Interest of $1 for any time 
within a year from any date. Also the EXACT diffarence in time with Interest 

at the above rates. 


EXPLANATION : 

By Rows, are meant the parallel lines of decimals running right and left. 

By Columns, are meant the vertical lines of decimals running up and down. 

The figures above the names of the months at the top of the columns, commencing at 0 and end- 
ing with the 11, denote time in months, from the beginning of the year; those on the extreme left 
of each row of decimals, commencing at 0 and ending with 29, denote time in days from the begin¬ 
ning of the month. 

The object of the Tables, is to furnish the Interest of One Dollar for any time, from any date, 
within a year. 

This interest multip’ied by the give number of dollars will evidently produce the required in¬ 
terest of any sum of money for any period of time within a year, frem the given date. 

When the time is known : Look Jor the given number of months at the top of the column . and 
pass down until opposite the number of days on the extreme left, and ice find the interest of one 
dollar for the given time, at the rate per cent, upon which the table has been constructed. 

For instance: Find the interest of one dollar for 7 months and 13 days at 7%- Turn to 7% 
Table. Find 8 months at the top, and pass down the column until opposite 13, and we find the 
decimal .043361, the required interest. 

If the time is over a year, multiply the given rate per cent, by the number of years, and add the 
product in cents to the decimal of the given months and days. 

For instance: The time is 3 years, 7 months and 13 days at 7%; we should say, 7X3=.21-f- 
.043361=.2 3361, %. e. 25c. 

We will now illustrate the greatest advantage these tables possess over all others, namely: Find¬ 
ing the difference in time between the date of signing and the maturity date, with the interest of 
one dollar, at the same moment. 

Thus: A note for 81.00 was signed March 3rd, and was paid November 26th the. same year. 
Required the interest @ 7%. 

Turn to 7% table. Over March is figure 2; therefore, 2 months and 13 days have elapsed since 
the beginning of the year. Each column represents a month of 30 days. Consequently, we count 
bach from the decimal of the maturity date 2 columns and 13 rows. The decimal of November 
20th, found as above is .063388; now, count back two columns, and we are at the decimal .051722, 
and, counting back (upwards) 13 rows, we find, opposite 7%, the required interest, to wit: .049194. 

If the difference in time consists of exactly 360 days, the eye will stop at the beginning of the 
table on the decimal .000000. In that case the interest equals as many cents as i3 represented by 
the per cent. Instead of counting back, as shown in the foregoing, if it is preferred, the same 
result may be obtained by deducting the smaller decimal from the larger, provided both dates are 
in the same year. The method of counting back is more rapid, and there is scarcely a chance for 
error. 

When a note is signed in one year and matures in another, there are not columns enough on the 
left of the maturity date to make up for the signing date. In that case, count back from the 
maturity date, as before, until you reach the 0 column; then trace the row to the extreme right 
hand column, and count back from it the remaining columns and rows. 

The last figure of every decimal in the tables is a repeater, i. e.: it can be annexed consecutively 
to any desired period of accuracy. 

The figures at the bottom of the columns are used when th exact time (i. e , 365 days to the 
year) is required. Thus, in passing over February, two days are deducted from the maturity date, 
unless it be a leap year, in which case only one day is deducted. The figures under the ether 
months should be added to the maturity date—as many as are found between the signing and 
maturity dates,—then proceed exactly as when counting thirty days to each month. 

It is apparent that these tables will save great labor and time in Partial Payments, Averaging 
Accounts, etc., as well as in computing interest. In multiplying the decimals, use the Perfected 
Method of Multiplication. 





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